K.V.P.Y 2016 question

$\begin{aligned} f(8) & = 15 \\ f(2 \times 4) & = 15 \\ f(2) + f(4) & = 15 \\ f(2) + f(2\times 2) & = 15 \\ f(2) + f(2) + f(2) & = 15 \\ 3f(2) & = 15 \\ \implies f(2) & = 5 \\ \implies f(4) & = f(2) + f(2) = 10 \end{aligned}$

$\implies f(48) = f(4\times 12) = f(4) + f(12) = 10 + 24 = \boxed{34}$

From $f(xy) = f(x) + f(y)$ and other two conditions, we get

$f(96) = f(48) + f(2) = f(12) + 3f(2) = 39$ . $...(1)$

Which implies

$f(2) = 5$ .

Putting this in $(1)$ , we get

$f(48) = 39 - 5 = \boxed{34}$ .

There are so many chains to reach value of $f(48)$ from given conditions. I have shown one of them.

Other solutions are always welcome.